Proof. The value y=1y1y=1 in the ultrametric triangle inequality gives the (*) as result. Secondly, let’s assume the condition (*). Let xxx and yyy be non-zero elements of the field KKK (if x⁢y=0xy0xy=0 then 3 is at once verified), and let e.g. |x|≦|y|xy|x|\leqq|y|. Then we get
|xy|=|x|⋅|y|-1≦1xynormal-⋅xsuperscripty11\displaystyle|\frac{x}{y}|=|x|\cdot|y|^{{-1}}\leqq 1, and thus according to (*),

Proof. Let e.g. |x|>|y|xy|x|>|y|. Surely |x+y|≦|x|xyx|x\!+\!y|\leqq|x|, but also |x|=|(x+y)-y|≦max⁡{|x+y|,|y|}xxyyxyy|x|=|(x\!+\!y)\!-\!y|\leqq\max\{|x\!+\!y|,\,|y|\}; this maximum is |x+y|xy|x\!+\!y| since otherwise one would have |x|≦|y|xy|x|\leqq|y|. Thus the result is: |x+y|=|x|xyx|x\!+\!y|=|x|.

Note. The metric defined by a non-archimedeanvaluation of the field KKK is the ultrametric of KKK. Theorem 2 implies, that every triangle of KKK with verticesAAA, BBB, CCC (∈KabsentK\in K) is isosceles: if |B-C|≠|C-A|BCCA|B\!-\!C|\neq|C\!-\!A|, then |A-B|=max⁡{|B-C|,|C-A|}ABBCCA|A\!-\!B|=\max\{|B\!-\!C|,\,|C\!-\!A|\}.

Theorem 3.

The valuation|⋅|:K→ℝfragmentsnormal-|normal-⋅normal-|normal-:Knormal-→R|\cdot|:K\to\mathbb{R} of the field KKK is archimedean if and only if the set

Proof. If |⋅|fragmentsnormal-|normal-⋅normal-||\cdot| is non-archimedean, then |n⋅1|=|1+…+1|≦max⁡{|1|}=1normal-⋅n11normal-…111|n\cdot 1|=|1\!+\ldots+\!1|\leqq\max\{|1|\}=1, and the multiples are bounded. Conversely, let
|n⋅1|<M⁢∀n∈ℤ+normal-⋅n1Mfor-allnsubscriptℤ|n\cdot 1|<M\,\,\forall n\in\mathbb{Z}_{+}. Now one obtains, when |x|≦1x1|x|\leqq 1:

or |x+1|<(n+1)⁢Mnx1nn1M|x\!+\!1|<\sqrt[n]{(n\!+\!1)M} for all nnn. As nnn tends to infinity, this nthsuperscriptnthn^{\mathrm{th}}root has the limit 1. Therefore one gets the limit inequality|x+1|≦1x11|x\!+\!1|\leqq 1, i.e. the valuation is non-archimedean.